Imaginary number conundrum, how does i*√(9*i) = -√(-9*i)?

Question

I asked this question before but I think it was unclear.
I mean show how like this, a case I understand:

i*√9 = √(-1)*√9 = √((-1)*9) = √(-9)

Start from the left, and go to get the result on the right. Here's my attempt, following the above:

i*√(9*i) = √(-1)*√(9*i) = √((-1)*(9*i)) = √(-9*i)

But this is wrong, somehow I should end up with -√(-9*i) but I don't see how. What is the missing step. Thanks.

Dr D. How can they both be correct? How can a number equal it's negative if the number is not zero?

Dr. D. I see you deleted that they are both correct, but I still get the negative sign with MathCad and Maple? I think there is a step in my attempt that I missed. Maybe it is a branch cut issue.

I see you show that they are equal by using the exponential forms, but can you do so using the method I attempted? Thanks.

Answer 1

Forget about the 9. Let's just show that
i*√i = -√(-i)

LHS:
i = exp(iπ/2)
√i = exp(iπ/4)
i*√i = exp(i*pi;/2) * exp(i*pi;/4)
= exp(i*3π/4)

RHS:
-i = exp(-i*π/2)
√(-i) = exp(-i*π/4)
-1 = exp(i*π)
-1 * √(-i) = exp(i*π)* exp(-i*π/4)
= exp(i*3π/4)
= LHS

*EDIT*
I just saw your comments. I changed my earlier answer after checking it with maple.

It is better to work with the exponential form of hte complex numbers because the discrepancy you're seeing is because of the man made definition of radicals.
EG. √(1) = √(-i^2) = √(-1) * √(i^2) = i * i = -1
The man made rule that says the radical is always +ve fails when you're dealing with complex numbers. It's a real number rule. It's better to work with exponentials than radicals when dealing with complex numbers.

I'll do some more reading on the radicals of complex numbers and get back to you. It may very well be a branch issue as you said.

*EDIT*
Note that in the RHS analysis above, I chose
-i = exp(-i*π/2)
But I could have also chosen
-i = exp(i*3π/2)
in which case the RHS would have been equal to the negative of the LHS, or the same result you got using radicals.

I am not sure how radicals are defined in complex numbers. To my knowledge they are multivalued. I am not aware of any convention.

*EDIT*
OK I found what I was looking for. The principal value of an argument is defined in the range:
(-π, π]
That's why we couldn't choose
-i = exp(i*3π/2)

So according to that convention, this line of your working is where the error occurred:
√(-1)*√(9*i) = √((-1)*(9*i))
The LHS has argument 3π/4, while the RHS has argument -π/4.
That's why there should be a -ve sign on the RHS, and that's why working with radicals is not the best way.

Answer 2

First I would dip the wand in the pixie dust. Then I would skip about magicking things. Like a magicking pixie fairy waving a wand. Tra la la la la

Answer 3

You're right. Answer provided, wrong.

Answer 4

Your calculations are correct.

i*sqrt(9i) = sqrt(-9i) not -sqrt(-9i)

The second one is equal to -i